Physics 1. Mechanics Problems with solutions 1 Coordinates and vectors. Problem 1.1. Find the distance between two points P 1 =(r, ϕ 1 ) and P 2 =(r, ϕ 2 ) (polar coordinates). Solution. s 2 =(x 2 - x 1 ) 2 +(y 2 - y 1 ) 2 =(r cos ϕ 2 - r cos ϕ 1 ) 2 +(r sin ϕ 2 - r sin ϕ 1 ) 2 =2r 2 (1 - cos ϕ 1 cos ϕ 2 - sin ϕ 1 sin ϕ 2 ) =2r 2 [1 - cos(ϕ 2 - ϕ 1 )] =4r 2 sin 2 ϕ 2 - ϕ 1 2 Problem 1.2. Find the distance between P 1 =(R 1 ,θ 1 ,ϕ 1 ) and P 2 =(R 2 ,θ 2 ,ϕ 2 ) (spherical coordinates). Solution. s 2 =(x 2 - x 1 ) 2 +(y 2 - y 1 ) 2 +(z 2 - z 1 ) 2 =(R 2 sin θ 2 cos ϕ 2 - R 1 sin θ 1 cos ϕ 1 ) 2 +(R 2 sin θ 2 sin ϕ 2 - R 1 sin θ 1 sin ϕ 1 ) 2 +(R 2 cos θ 2 - R 1 cos θ 1 ) 2 = R 2 1 + R 2 2 - 2R 1 R 2 cos θ 1 cos θ 2 - 2R 1 R 2 sin θ 1 sin θ 2 cos(ϕ 2 - ϕ 1 ) Problem 1.3. We define elliptical coordinates as follows: r = x 2 /a 2 + y 2 /b 2 and ϕ as in polar coordinates. Find x, y as functions of r, ϕ. Solution. y = x tan ϕ → 1
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Physics 1. Mechanics
Problems with solutions
1 Coordinates and vectors.
Problem 1.1. Find the distance between two points P1 = (r, ϕ1) and P2 = (r, ϕ2) (polar
coordinates).
Solution.
s2 = (x2 − x1)2 + (y2 − y1)
2
= (r cos ϕ2 − r cos ϕ1)2 + (r sin ϕ2 − r sin ϕ1)
2
= 2r2(1− cos ϕ1 cos ϕ2 − sin ϕ1 sin ϕ2)
= 2r2[1− cos(ϕ2 − ϕ1)]
= 4r2 sin2
(ϕ2 − ϕ1
2
)
Problem 1.2. Find the distance between P1 = (R1, θ1, ϕ1) and P2 = (R2, θ2, ϕ2) (spherical
coordinates).
Solution.
s2 = (x2 − x1)2 + (y2 − y1)
2 + (z2 − z1)2
= (R2 sin θ2 cos ϕ2 −R1 sin θ1 cos ϕ1)2
+ (R2 sin θ2 sin ϕ2 −R1 sin θ1 sin ϕ1)2 + (R2 cos θ2 −R1 cos θ1)
2
= R21 + R2
2 − 2R1R2 cos θ1 cos θ2 − 2R1R2 sin θ1 sin θ2 cos(ϕ2 − ϕ1)
Problem 1.3. We define elliptical coordinates as follows: r =√
x2/a2 + y2/b2 and ϕ as in polar
coordinates. Find x, y as functions of r, ϕ.
Solution.
y = x tan ϕ →
1
Physics 1. Mechanics Problems
x2
a2+
x2 tan2 ϕ
b2= r2 →
x = abr cos ϕ(a2 sin2 ϕ + b2 cos2 ϕ)−1/2
y = abr sin ϕ(a2 sin2 ϕ + b2 cos2 ϕ)−1/2
Problem 1.4. In elliptical coordinates (r =√
x2/a2 + y2/b2 and ϕ as in polar coordinates) find
distance between two points P1 = (r, ϕ1) and P2 = (r, ϕ2).
Solution.
s2 = (x2 − x1)2 + (y2 − y1)
2
= (abr)2
(cos ϕ2√
a2 cos2 ϕ2 + b2 sin2 ϕ2
− cos ϕ1√a2 cos2 ϕ1 + b2 sin2 ϕ1
)2
+ (abr)2
(sin ϕ2√
a2 cos2 ϕ2 + b2 sin2 ϕ2
− sin ϕ1√a2 cos2 ϕ1 + b2 sin2 ϕ1
)2
=(abr)2
a2 cos2 ϕ1 + b2 sin2 ϕ1
+(abr)2
a2 cos2 ϕ2 + b2 sin2 ϕ2
− 2(abr)2 cos(ϕ2 − ϕ1)√
a2 cos2 ϕ1 + b2 sin2 ϕ1
√a2 cos2 ϕ2 + b2 sin2 ϕ2
Problem 1.5. Two different coordinate systems are established on a straight line, x and x′, which
are related as follows: x/a = (x′/b)3, where a and b are constants. The distance element in terms of
x is given by ds2 = dx2. Find the expression for the distance element in terms of x′.
Solution.
ds2 = dx2 = (dx
dx′)2dx′
2= (3ax′2/b3)2dx′
2
Problem 1.6. Two different coordinate systems are established on a plane, (x, y) and (x′, y′),
which are related as follows:
x′ = x cos θ − y sin θ, y′ = x sin θ + y cos θ,
where θ = const. Find the distance element in terms of coordinates (x′, y′), if (x, y) are Cartesian.
Solution. Inverting (solving for x and y) we get
x = x′ cos θ + y′ sin θ, y = −x′ sin θ + y′ cos θ
2
Physics 1. Mechanics Problems
Now
ds2 = dx2 + dy2 = (∂x
∂x′dx′ +
∂x
∂y′dy′)2 + (
∂y
∂x′dx′ +
∂y
∂y′dy′)2
= (cos θdx′ + sin θdy′)2 + (− sin θdx′ + cos θdy′)2 = dx′2+ dy′
2
Problem 1.7. Same as above but the relation reads
x′ = x cos θ + y sin θ, y′ = x sin θ + y cos θ,
.
Solution. Inverting one gets
x =x′ cos θ − y′ sin θ
cos2 θ − sin2 θ
y =−x′ sin θ + y′ cos θ
cos2 θ − sin2 θ
Now (cos2 θ − sin2 θ = cos 2θ)
ds2 = dx2 + dy2 = (∂x
∂x′dx′ +
∂x
∂y′dy′)2 + (
∂y
∂x′dx′ +
∂y
∂y′dy′)2
=1
cos2 2θ[(cos θdx′ − sin θdy′)2 + (− sin θdx′ + cos θdy′)2]
=1
cos2 2θ[dx′
2+ dy′
2 − 2 sin 2θdx′dy′]
Problem 1.8. New coordinates (ρ′, ϕ′) are introduced which are related to the ordinary polar
coordinates (ρ, ϕ) as follows: ϕ′ = ϕ, ρ′ = 1/ρ. Find the distance element.
Solution.
ϕ′ = ϕ, ρ′ = 1/ρ → ϕ = ϕ′, ρ = 1/ρ′
ds2 = dρ2 + ρ2dϕ2 = (∂ρ
∂ρ′dρ′ +
∂ρ
∂ϕ′dϕ′)2 + (1/ρ′)2(∂ϕ
∂dρ′dρ′ +
∂ϕ
∂ϕ′dϕ′)2
= (1/ρ′)4dρ′2+ (1/ρ′)2dϕ′2
Problem 1.9. Starting with cartesian coordinates we define new ones as follows: x′ = ax + by,
y′ = cx + dy, where a, b, c, d are some constant parameters. What conditions on these parameters
should be satisfied in order that the new coordinates also be cartesian and the measure of the distance
Problem 6.2. A particle is in the stable equilibrium in the potential energy U(x) = U0[1 −l2/(l2 +x2)]. Suddenly it gets a small addition of energy E ′. Assuming that the oscillations are small
find the frequency and amplitude.
Problem 6.3. A particle moves in a well of the shape y = ax2 without friction (potential energy
U = mgy). Show that the motion can be described as a harmonic oscillation and find the frequency.
Solution. Potential energy U = mgy = mgax2. Kinetic energy
K = (m/2)(x2 + y2) =m
2(1 + 4a2x2)x2
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Physics 1. Mechanics Problems
When ax 1 we have K ≈ m2x2 and ω2 = 2ga.
Problem 6.4. A body with the mass m is attached to a spring (spring constant k). The other
end of the spring is brought into the motion according to the law x = x0 cos(ωt). The friction acting
on the body is Fx = −bvx. Show that the body can oscillate with a constant amplitude and find this
amplitude.
Solution. Let y be the coordinate of the body. Then F = −k∆l = −k(y − x). We have
my = −k(y − x)− by ⇒ my + by + ky = kx0 cos(ωt)
Therefore, y = A cos(ωt− φ). Substitute and complete.
Problem 6.5. Find the motion of an oscillator with the natural frequency ω0 and mass m under
the force F = F0 + F1 cos(ωt), ω 6= ω0.
Solution. If F = F0 the solution is x1 = F0/mω2. When F = F1 cos(ωt) the solution is
x2 = A cos(ωt− φ). The total solutions is x1 + x2.
Problem 6.6. Find the frequency of small radial oscillations of a particle with a mass m near a
circular orbit r = r0 in a central potential U(r) = −k/r.
Tangential acceleration is parallel to the velocity vector, so that a‖ = a · v/|v|:
a‖ = (axvx + ayvy)/(v2x + v2
y)1/2
= −γ√
γ2 + ω2A exp(−γt)
Problem 10.6. A particle of the mass m moves in the potential energy U(x) = −Ax2/2+Bx3/3,
A > 0, B > 0. Find the frequency of small oscillations.
Solution. Equilibrium is where Fx = −(dU/dx) = 0:
−dU
dx= −Ax + Bx2 = 0 → x1 = 0, x2 = A/B.
Whether the equilibrium is stable or not is determined by the second derivative (d2U/dx2) = −A +
2Bx. If (d2U/dx2) < 0 the point is a maximum and the equilibrium is unstable, if (d2U/dx2) > 0
the point is a minimum and the equilibrium is stable. In x = x1 = 0 we have (d2U/dx2) = −A < 0
so that this is an unstable equilibrium. In x = x2 = A/B we have (d2U/dx2) = A > 0 and this is a
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Physics 1. Mechanics Problems
stable equilibrium.
Near the equilibrium point the potential energy can be Tayor expanded:
U(x) ≈ U(x2) + 12A(x− x2)
2.
Let us denote X = x− x2 then vx = x = X and the energy conservation is written as
12mX2 + 1
2AX2 = const
which immediately gives (according to the general rule)
ω2 = A/m.
Problem 10.7. A hollow cylinder is sliding without friction (no rolling) with the velocity v. The
cylinder comes to a surface with friction. What is the final velocity of the cylinder ?
Solution. When the cylinder comes to the surface with friction it is decelerated by the friction
force and at the same time its rotation is acceletrated until the cylinder begins to roll without sliding.
In the rolling state the friction force is zero and the velocity does not change. Let the friction force
magnitude be Fs. Then the deceleration is
mV = −Fs → V (t) = v − Fs
mt,
whereV (t) is the velocity of the center-of-mass in the moment t. The moment of the friction force
(torque) accelerates the rotation around the axis passing through the center-of-mass, as follows:
Iω = Fsr → ω(t) =Fsr
It,
where ω(t) is the angular velocity of rotation and I is the moment of inertia around the axis passing
through the center-of-mass. According to the definition I =∑
mir2i , where ri is the distance from
the rotating mass (small part of the body) from the rotation axis. All points of the cylinder are at
the same distance r from the rotation axis, so that we have
I =∑
mir2i =
∑i
mir2 = (
∑i
mi)r2 = mr2.
From (??) and (??) we find
ω(t) =Fs
mrt.
The velocity and angular velocity stop changing when sliding stops. The condition of the rolling
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Physics 1. Mechanics Problems
without sliding is V = ωr which gives
v − Fs
mt =
Fs
mrtr =
Fs
mt,
and therefore, in the moment when the velocity stops changing
Fs
mrt = v/2.
Substituting into (??) we have
Vfinal = v/2.
Problem 10.8. A rod of the mass M and length l can rotate in the vertical plane around the
axis which passes through the point at a < l/2 from its upper end. A bullet of the mass m M
with the horizontally directed velocity v strikes the rod at the upper end and remains in it. What is
the rotation angle of the rod ?
v
m, v
E ! mc2
λ
µ
µ " m " M M m
R
2
Solution. Angular momentum is conserved during the collision, which means
Iω = mva → ω = mva/I,
where ω is the angular velocity of the rotation just after the collision, and I is the moment of inertia.
After the collision the energy is conserved, that is, the initial kinetic energy goes into the potential
energy as the bar rotates to the angle θ and its center-of-mass rises. Therefore, for the maximum
angle we have
Iω2/2 = MgL(1− cos θ) → cos θ = 1− Iω2/2MgL.
where L = l/2 − a is the distance between the axis and center-of-mass of the bar (we neglect the
mass of the bullet).
Calculation of I: Let us choose coordinate origin in the axis, and the positive direction downwards.
41
Physics 1. Mechanics Problems
Then
I =
∫ l−a
−a
x2(M/l)dx =M
3l[(l − a)3 + a3]
Combine all calculations.
Problem 10.9. An electron moving with the energy E mc2 toward the coordinate origin emits
a photon with the wavelength λ forward. What wavelength measures a non-moving observer in the
coordinate origin ?
Solution. We shall use directly the expression for the Doppler effect
λ′
λ=
1
γ(1− v cos θ/c).
In our case the source is moving towards the receptor, so that θ = 180 and
λ′
λ=
√1− v/c
1 + v/c.
We need velocity which can be found as follows: γ = E/mc2 and v = c√
1− 1/γ2.
Problem 10.10. Potential energy (2D) is given by U = A cos ϕ/ρ2. A particle of the mass m is
in the point r = (a, a) and its velocity is v ⊥ r. Find the normal acceleration.
Solution. By definition an ⊥ v. Since r ⊥ v we have an ‖ r. Thus, the magnitude an = |a · r|.a) The long way. r = (ax + ay)/
√2a2 = (x + y)/
√2. Express the potential energy in Cartesian
coordinates x = r cos ϕ, y = sin ϕ: U = Ax(x2 + y2)−3/2. Then
F = −∂U
∂xx− ∂U
∂yy
= −(
A
r3− 3Ax2
r5
)x +
(3Axy
r5
)y
=A(3 cos2 ϕ− 1)
r3x +
3A sin ϕ cos ϕ
r3y.
Now a = F /m, r = a√
2, and ϕ = 45, so that an = A/2ma3.
b) The short way. F = −(∂U/∂r)r−(1/r)(∂U/∂ϕ)ϕ, so that an = |(∂U/∂r)|/m = |2A cos ϕ/r3|/m =
A/2ma3.
Problem 10.11. A satellite of the mass m is moving on an elliptical orbit round the Earth (mass
M m), so that rmax = 2rmin. The satellite energy E < 0 is known. Find the angular momentum.
Solution. When r = rmin or r = rmax the radial velocity vanishes r = 0, so that in these points
42
Physics 1. Mechanics Problems
the energy and momentum conservation give
E =J2
2mr2− GMm
r.
Thus,
r1,2 =GMm±
√(GMm)2 − 2|E|J2/m
2|E|
(here we used E = −|E| < 0) and
GMm +√
(GMm)2 − 2|E|J2/m
GMm−√
(GMm)2 − 2|E|J2/m= 2.
Solving this equation we get
J =2GMm
√m
3√|E|
.
Another way of solving: from the energy expression we have
J2/2m = r2E + GMmr = (2r)2E + GMm(2r),
so that r = −GMm/3E and
J =√
2m(r2E + GMmr)
same as above.
Problem 10.12. Two disks of the masses m1 and m2 and radii R1 and R2 are connected to a
massless rod which can rotate in the vertical plane. The disk centers are at the distances l1 and l2
from the rotation axis. Find the frequency of small oscillations.
m1, R1
m2, R2
l1
l2
m R M
h r
B = (0, B, 0) x − y
q > 0 m
2
43
Physics 1. Mechanics Problems
Solution. If the system moved to the angle θ 1 and the angular velocity is θ, the energy is
E =Iθ2
2+ mgL(1− cos θ) =
Iθ2
2+
mgLθ2
2,
where
I = (m1R
21
2+ m1l
21) + (
m2R22
2+ m2l
22)
is the moment of inertia relative to the axis, and
L =m1l1 −m2l2
m1 + m2
is the distance of the center of mass from the axis. The frequency is ω2 = mgL/I.
Problem 10.13. To a disk of the mass M and radius R a smaller disk of the mass mv and radius
r is connected. At what height a horisontal force should be applied in order that the body start (in
the very first moment) to roll without sliding. There is no friction.
m1, R1
m2, R2
l1
l2
m R M
h r
B = (0, B, 0) x − y
q > 0 m
2
Solution. Let the force be F . The acceleration of the center of mass is a = F/(M + m). The
angular acceleration around the contact point is α = N/I = Fh/I, where the moment of inertia is
I = (MR2/2 + MR2) + (mr2/2 + mr2) = (3/2)(MR2 + mr2). If there is no sliding, the linear and
angular acceleration are related by a = αL, where L = (MR + mr)/(M + m) is the distance of the
center of mass from the rotation axis (contact point). Thus, we have F/(M + m) = (Fh/I)L, so
that h = I/L(M + m).
Problem 10.14. There is a homogeneous magnetic field B = (0, B, 0) between two plates,
parallel to the x − y plane. A particle of the mass m and electric charge q > 0 enters the space
between the planes through the lower plate, when its velocity is v = (v cos 45, 0, v sin 45). What
is the minimal distance between the plates for which the particle will come back to the lower plate
44
Physics 1. Mechanics Problems
(without touching the upper one) ? Gravity is negligible.
v = (v cos 45, 0, v sin 45)
x
z
v
v1 = (0.5c, 0.5c, 0)
|vrel| v2 = (−0.5c, 0.5c, 0)
R M
3
Solution. The particle moves on a circle of the radius r which is obtained from F = qvB =
mv2/r, that is, r = mv/qB, as shown in the figure.
Physics 1. Mechanics Solutions 2
x
z
v
From the figure it is clear that the distance between the two planes should be L > r + r/√
2.
Problem 1.6. The easiest way is to rotate the coordinates so that the new x axis will be along v1,
then the new y axis will be along v2. In the new coordinates v1 = (0.5√
2c, 0, 0), v2 = (0, 0.5√
2c, 0).
Taking v1 ≡ v0 as the velocity of the moving frame (S ′), and v2 as the velocity of a body we are
looking at, we have
v′
x =vx − v0
1 − vxv0/c2= −v0 = −0.5
√2c, (11)
v′
y =vy
γ(1 − vxv0/c2)=
vy
γ=
0.5√
2c
γ. (12)
where we used vx = (v2)x = 0, and
γ = (1 − v2
1/c2)−1/2 =
√2. (13)
Thus, we have v′
x = −0.5√
2c, v′
y = 0.5c, and v′ =√
v′
x2 + v′
y2 = 0.5
√3c.
For advanced only !
Problem 1.7. Let us consider a small part of the mass of the length dr and cross section dA at
the radius r. This part has a mass dm = ρdAdr, where the density ρ = 3M/4πR3. Is is attracted to
the center by the mass M ′ = ρ(4πr3/3) inside the radius r. This force is balanced by the pressure
forces. From the inside the pressure pushed it outwards with the force p[r]dA. From the outside the
pressure pushes it inward with the force p[r + dr]dA. Thus, we have
p[r]dA − p[r + dr]dA =GM ′dm
r2=
GMrρ
R3dAdr. (14)
3
From the figure it is clear that the distance between the two planes should be L > r + r/√
2.
Problem 10.15. An observer on Earth sees one galaxy moving with the velocity v1 = (0.5c, 0.5c, 0)
and another with v2 = (−0.5c, 0.5c, 0). Find their relative velocity |vrel|.Solution. The easiest way is to rotate the coordinates so that the new x axis will be along v1,
then the new y axis will be along v2. In the new coordinates v1 = (0.5√
2c, 0, 0), v2 = (0, 0.5√
2c, 0).
Taking v1 ≡ v0 as the velocity of the moving frame (S ′), and v2 as the velocity of a body we are
looking at, we have
v′x =vx − v0
1− vxv0/c2= −v0 = −0.5
√2c,
v′y =vy
γ(1− vxv0/c2)=
vy
γ=
0.5√
2c
γ.
where we used vx = (v2)x = 0, and
γ = (1− v21/c
2)−1/2 =√
2.
Thus, we have v′x = −0.5√
2c, v′y = 0.5c, and v′ =√
v′x2 + v′y
2 = 0.5√
3c.
Problem 10.16. Potential energy is given as U = −A/r4, A > 0. When the particle (mass m)
is in the point (a, 0, 0) its velocity is v = (0, v, 0) and its energy is negative. What is the maximum
45
Physics 1. Mechanics Problems
distance between the particle and the coordinate origin ?
Solution. Central force means that J = mr×v = const and E = mv2r/2+J2/2mr2+U = const.
From the initial conditions we have J = J z, J = mav, E = mv2/2 − A/a4. Since E < 0 we have
v2 < 2A/ma4. In the closest and farthest points vr = 0 so that we have
J2
2mr2− A
r4= E
where J , m, E < 0, and A are known. The equation is easily solved to give
r2 = − J2
4m|E|±
√(
J2
4m|E|)2 +
A
|E|
The solution with − is not physical (r2 < 0) which means that there is no minimum radius: the
particle reaches the maximum radius, bounces back and falls into r = 0.
Problem 10.17. A particle moves according to x = r0 cos(ωt), y = r0 sin(ωt), z = kt2/2. Find
the normal acceleration.
Solution.
v = ωr0(− sin(ωt)x + cos(ωt)y) + ktz
a = −ω2r0(cos(ωt)x + sin(ωt)y) + kz
an = |a× v|
= ωr0
√ω2k2t2 + k2 + ω4r2
0√ω2r2
0 + k2t2
Problem 10.18. Potential energy is U(x) = A(x − a)2(x − b)2, A > 0. Find the frequency of
small oscillations.
Solution. Equilibrium: dU/dx = 0 → x = a or x − b. It is easy to find that U = A(b − a)2X2
where X = x− a or X = x− b. Thus, ω2 = 2A(b− a)2/m.
Problem 10.19. Three identical cylinders are in equilibrium (see figure). Find the minimum
friction coefficient between the cylinders.
dve‘zd z‘ ‘vn . z = at2/2 , y = r0 sin(ωt) , x = r0 cos(ωt) itl rp wiwlg